What is the center of the circle given by the equation (x-3) 2 + (y-9) 2 = 16?

A. (-9,-3)

B. (-3,-9)

C. (3,9)

D. (9,3)

The center is (3,9) and the radius is 4

Step-by-step explanation:

We can write the equation for a circle as

(x-h) ^2 + (y-k) ^2 = r^2

Where (h, k) is the center and r is the radius

(x-3) ^2 + (y-9) ^2 = 4^2

The center is (3,9) and the radius is 4

C

Step-by-step explanation:

(x-a) ^2 + (y-b) ^2 = r^2

(a, b) = Center

r = radius

SO ...

That means the center of the circle " (x-3) ^2 + (y-9) ^2 = 16" is ...

(3,9) (C)